Question

3. Resolver em R a inequação (4x + 1)(5x + 3) > 0

Answer 1

Resposta:

(4x+1)(5x+3) > 0

20$x^{2}$+12x+5x+3 > 0

$20x^{2}$ +17x+3

$x=\frac{-b\sqrt{b^{2}-4ac } }{2a}$

$x= \frac{-17+-\sqrt{17^{2} -4*20*3} }{2*20}$

$x=\frac{-17+-\sqrt{49} }{80}$

$x=\frac{-17+-7}{80}$

$x'=\frac{-17+7}{80} = \frac{-10}{80} =-\frac{1}{8}$

$x"=\frac{-17-7}{80} =\frac{-24}{80} = -\frac{3}{10}$

$(x'=-\frac{1}{8} , x"=-\frac{3}{10} ) > 0$

Espero ter ajudado :)